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A vessel is floating in dockwater of RD 1.005with her starboard WNA mark 30mm below and her port WNA mark 60mm below the waterline . If her summer SW draught is 8.4m, TPC is 30 and FWA is 160 mm, calculate how much cargo can be loaded to bring the vessel to her vessel draught in SW.
Solution:
Dockwater RD = 1.005,
Stbd WNA mark = 30mm below
Port WNA mark = 60mm below
Mean draft = (30 + 60) / 2
= 45mm
=4.5cm,below the water line.
Distance from WNA to water(W) = 50mm
= 5cm
So, sinkage available = (50 – 45)
= 5mm
= 0.5cm
We know that :
Water (W) is 1/48 of summer draft
= (1/48 x 8.4)
= 0.175m
= 17.5cm
Sinkage = (17.5 + 0.5 )
= 18cm
Total sinkage = (18 + 12.8)
= 30.8cm
As we know:
Change in draft =
(Change in RD )x(FWA)
0.025
= (1.025 – 1.005 )x 16 / 0.025
= 12.8cm
Again, TPC = (A/100) x density
30 = (A/100) x 1.025
A = 30/ 1.025
A = 29.268 m2
Now , TPC for density of 1.005
TPC = (A/100) x 1.005
= (29.268 x 1.005)
= 29.41 t/cm.
Hence ,cargo can be loaded = (30.8 x 29.41)
= 905.82 t
A ship loads in fresh water to her salt water marks and proceeds along a
river to a second port consuming 20 tonnes of bunkers. At the second port,
where the density is 1016 kg per cu. m, after 120 tonnes of cargo have been
loaded, the ship is again at the load salt water marks. Find the ship’s load
displacement in salt water.
Joe Bee. Can you please send me the solved question of the above question you posted. please
A mariner Enters port with 17,700 tons displacement,KG=28 feet, and GG=1.05 feet.A piece of Heavy machinery which weighs 180 tons is lifted by shore based cranes and placed on death so that its centre of gravity is 47 ft above the ship Khel and 19 feet two star board of centerline what will be the final mean draught gvm and angle of list when loading is complete?